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We can't use Green's Theorem directly, because the integrand is undefined at the origin, which is inside C.

If you try to do this without Green's Theorem, parameterizing C by

x = 2 cos t, y = 3 sin t for t in [0, 2π] yields

∫(t = 0 to 2π) <-3 sin t/(4(4 cos² t + 9 sin² t), 2 cos t/(4(4 cos² t + 9 sin² t)> · <-2 sin t, 3 cos t> dt

= ∫(t = 0 to 2π) (3/2) dt/(4 cos^2(t) + 9 sin^2(t), which is not a pleasant integral at all.

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Try this instead:

Let C* be the contours C (as above; counterclockwise), and C', the unit circle x^2 + y^2 = 1 with clockwise orientation. Note that the region between C and C' avoids the origin; so Green's Theorem is applicable.

On one hand, ∫c* F · dr

= ∫c* [(-y/(4x^2 + 4y^2)) dx + (x/(4x^2 + 4y^2)) dy]

= (1/4) ∫c [(-y/(x^2 + y^2)) dx + (x/(x^2 + y^2)) dy]

= (1/4) ∫∫ [(∂/∂x) x/(x^2 + y^2) - (∂/∂y) -y/(x^2 + y^2)] dA, by Green's Theorem

= (1/4) ∫∫ [(y^2 - x^2)/(x^2 + y^2)^2 + (x^2 - y^2)/(x^2 + y^2)^2] dA

= 0.

On the other hand, ∫c* F · dr = ∫c F · dr + ∫c' F · dr.

Hence, ∫c F · dr + ∫c' F · dr = 0

==> ∫c F · dr = -∫c' F · dr.

Due to orientation, -∫c' F · dr = ∫(-c') F · dr, where (-C)' is the contour about the unit circle with counterclockwise orientation.

This is a much easier integral to evaluate; use x = cos t, y = sin t for t in [0, 2π].

Finally, ∫c F · dr

= ∫(-c') F · dr

= ∫(t = 0 to 2π) <-sin t / 4, cos t / 4> · <-sin t, cos t> dt

= (1/4) ∫(t = 0 to 2π) dt

= π/2.

I hope this helps!

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